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09 Dec 2021, 05:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
33% (02:38) correct
67%
(02:52)
wrong
based on 81
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The sum of the digits of a four digit number is 32. What is the probability that such a number is divisible by 11?
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
Are You Up For the Challenge: 700 Level Questions
09 Dec 2021, 05:38
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Bunuel
The sum of the digits of a four digit number is 32. What is the probability that such a number is divisible by 11?
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
For Sum of four digits = 32
The set of digits should be either {8, 8, 8, 8} or {7, 7, 9, 9} or {7, 8, 8, 9}
Total Numbers formed by {8, 8, 8, 8} = 1 and that will be divisible by 11
Total Numbers formed by {7, 7, 9, 9} \(= \frac{4!}{2!2!} = 6\) and Four of them {7997, 7799, 9779, 9977} will be divisible by 11
Total Numbers formed by {6, 8, 9, 9} \(= \frac{4!}{2!} = 12\) and none of them will be divisible by 11
Total Numbers formed by {7, 8, 8, 9} \(= \frac{4!}{2!} = 12\) and Four of them {8789, 7898, 9878, 8987) will be divisible by 11
Total Numbers formed by {5, 9, 9, 9} \(= \frac{4!}{3!} = 4\) and None of them will be divisible by 11
Total Outcomes = 1+6+12+12+4 = 35
Total Numbers divisible by 11 = 1+4+4 = 9
Probability \(= \frac{9}{35}\)
ANswer: Option A
08 Oct 2024, 22:01
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Why can't the digits be {5,9,9,9} or {6,9,9,8}
For Sum of four digits = 32
The set of digits should be either {8, 8, 8, 8} or {7, 7, 9, 9} or {7, 8, 8, 9}
Total Numbers formed by {8, 8, 8, 8} = 1 and that will be divisible by 11
Total Numbers formed by {7, 7, 9, 9} \(= \frac{4!}{2!2!} = 6\) and Four of them {7997, 7799, 9779, 9977} will be divisible by 11
Total Numbers formed by {6, 8, 9, 9} \(= \frac{4!}{2!} = 12\) and none of them will be divisible by 11
Total Numbers formed by {7, 8, 8, 9} \(= \frac{4!}{2!} = 12\) and Four of them {8789, 7898, 9878, 8987) will be divisible by 11
Total Numbers formed by {5, 9, 9, 9} \(= \frac{4!}{3!} = 4\) and None of them will be divisible by 11
Total Outcomes = 1+6+12+12+4 = 35
Total Numbers divisible by 11 = 1+4+4 = 9
Probability \(= \frac{9}{35}\)
ANswer: Option A
GMATinsight
Bunuel
The sum of the digits of a four digit number is 32. What is the probability that such a number is divisible by 11?
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
A. \(\frac{9}{35}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{7}\)
D. \(\frac{4}{11}\)
E. \(\frac{2}{9}\)
For Sum of four digits = 32
The set of digits should be either {8, 8, 8, 8} or {7, 7, 9, 9} or {7, 8, 8, 9}
Total Numbers formed by {8, 8, 8, 8} = 1 and that will be divisible by 11
Total Numbers formed by {7, 7, 9, 9} \(= \frac{4!}{2!2!} = 6\) and Four of them {7997, 7799, 9779, 9977} will be divisible by 11
Total Numbers formed by {6, 8, 9, 9} \(= \frac{4!}{2!} = 12\) and none of them will be divisible by 11
Total Numbers formed by {7, 8, 8, 9} \(= \frac{4!}{2!} = 12\) and Four of them {8789, 7898, 9878, 8987) will be divisible by 11
Total Numbers formed by {5, 9, 9, 9} \(= \frac{4!}{3!} = 4\) and None of them will be divisible by 11
Total Outcomes = 1+6+12+12+4 = 35
Total Numbers divisible by 11 = 1+4+4 = 9
Probability \(= \frac{9}{35}\)
ANswer: Option A
